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Sagot :
To solve this problem, we will use the given formula for kinetic energy:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where:
- [tex]\( KE \)[/tex] is the kinetic energy,
- [tex]\( m \)[/tex] is the mass of the bottle,
- [tex]\( v \)[/tex] is the speed of the bottle.
We are given that the speed [tex]\( v \)[/tex] of the soda bottle is [tex]\( 4 \, \text{m/s} \)[/tex], and we need to calculate the kinetic energy for various masses.
### Step-by-Step Calculations
1. When the mass of the bottle is [tex]\( 0.125 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.125 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.125 \, \times 16 \][/tex]
[tex]\[ KE = 1.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
2. When the mass of the bottle is [tex]\( 0.250 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.250 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.250 \, \times 16 \][/tex]
[tex]\[ KE = 2.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
3. When the mass of the bottle is [tex]\( 0.375 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.375 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.375 \, \times 16 \][/tex]
[tex]\[ KE = 3.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
4. When the mass of the bottle is [tex]\( 0.500 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.500 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.500 \, \times 16 \][/tex]
[tex]\[ KE = 4.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
### Results Summary
- When the mass of the bottle is [tex]\( 0.125 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 1.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
- When the mass of the bottle is [tex]\( 0.250 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 2.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
- When the mass of the bottle is [tex]\( 0.375 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 3.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
- When the mass of the bottle is [tex]\( 0.500 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 4.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
By these calculations, we can observe how the kinetic energy of the bottle changes as its mass increases.
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where:
- [tex]\( KE \)[/tex] is the kinetic energy,
- [tex]\( m \)[/tex] is the mass of the bottle,
- [tex]\( v \)[/tex] is the speed of the bottle.
We are given that the speed [tex]\( v \)[/tex] of the soda bottle is [tex]\( 4 \, \text{m/s} \)[/tex], and we need to calculate the kinetic energy for various masses.
### Step-by-Step Calculations
1. When the mass of the bottle is [tex]\( 0.125 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.125 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.125 \, \times 16 \][/tex]
[tex]\[ KE = 1.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
2. When the mass of the bottle is [tex]\( 0.250 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.250 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.250 \, \times 16 \][/tex]
[tex]\[ KE = 2.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
3. When the mass of the bottle is [tex]\( 0.375 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.375 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.375 \, \times 16 \][/tex]
[tex]\[ KE = 3.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
4. When the mass of the bottle is [tex]\( 0.500 \, \text{kg} \)[/tex]:
[tex]\[ KE = \frac{1}{2} \times 0.500 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.500 \, \times 16 \][/tex]
[tex]\[ KE = 4.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \][/tex]
### Results Summary
- When the mass of the bottle is [tex]\( 0.125 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 1.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
- When the mass of the bottle is [tex]\( 0.250 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 2.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
- When the mass of the bottle is [tex]\( 0.375 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 3.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
- When the mass of the bottle is [tex]\( 0.500 \, \text{kg} \)[/tex], the [tex]\( KE \)[/tex] is [tex]\( 4.0 \, \text{kg} \cdot \text{m}^2 / \text{s}^2 \)[/tex].
By these calculations, we can observe how the kinetic energy of the bottle changes as its mass increases.
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